Answer to Question #209499 in Physics for Nyola

Question

A 50 capacitor , a 0.3 H inductor and an 80 Ω resistor are connected in series with a 120 V, 60 Hz power source. 

(1)What is the impedance of the circuit?

(2)How much current flows in the circuit?

What is the power factor?(2 marks)

(3)How much power is dissipated by the circuit?

(4)What must be the minimum rating in volt-amperes of the power source?

Answer 1

Assume the capacitance is 50 uC.

(1) The impedance:

$$Z=\sqrt{R^2+(X_L-X_C)^2},\\\space\\ Z=\sqrt{R^2+\bigg(2\pi f L-\frac{1}{2\pi f C}\bigg)^2}=2805\Omega.$$

(2) The current:

$$I=\frac{V}{Z}=0.0428\text{ A}.$$

The power factor:

$$\cos\phi=\frac RZ=0.0285.$$

(3) The power:

$$P=I^2R=3.42\text{ W}.$$

(4) The minimum rating in volt-amperes of the power source is

$$S=\frac{P}{\cos\phi}=120\text{ VA}.$$